Question

calculate the intensity of electric field at the point P due to the charges + 5 microcoulomb and minus 3.6 microcoulomb in the given diagram.
plzz solve it...​

Answer 1

Answer:

Electric Charge and Electric Field Example Problems with Solutions

Course:

Physics 2 (PH 202L)

Answer 2

Answer:

5 × 10⁻⁵ N/C.

Explanation:

The given triangle is the 3:4:5 configuration right triangle (as shown in the figure). Hence, the given θ will be equal to 37°.

Now, electric field intensity (E₁) due to +5 μC

E₁ = 9 × 10⁹ × $\frac{5~*~10^{-6}}{10~*~10^{-2}}$  = 5 × 10⁻⁵ N/C

And, electric field intensity (E₁) due to -3.6 μC

E₂ = 9 × 10⁹ × $\frac{3.6~*~10^{-6}}{6~*~10^{-2}}$ = 6 × 10⁻⁵ N/C

Now, consider the point P, you'll also get a similar right triangle of 3:4:5 configuration. By solving this, you can easily get your answer to be 5 × 10⁻⁵ N/C